# Cyclic Groups

Recall from the Generated Subgroups page that if $(G, \cdot)$ is a group then a generated subgroup of the elements $a_1, a_2, ..., a_n \in G$ denoted $(\langle a_1, a_2, ..., a_n \rangle, \cdot)$ is a group of all possible products (with respect to $\cdot$) of the elements $a_1, a_2, ..., a_n$ and $a_1^{-1}, a_2^{-1}, ..., a_n^{-1}$.

We will now look a specific type of group of interested known as cyclic groups.

Definition: A group $(G, \cdot)$ is said to be a Cyclic Group if there exists an element $a \in G$ such that $G = \langle a \rangle = \{ a^n : n \in \mathbb{Z} \}$, that is, all possible products of $a$ and $a^{-1}$ (with respect to the operation $\cdot$) generate all elements in $G$. The element $a$ for which $G = \langle a \rangle$ is called a Generator for the group $G$. |

For example, consider the group of integers modulo $n$, $(\mathbb{Z}_n, \cdot)$ where we recall that $\mathbb{Z}_n = \{ 0, 1, 2, ..., n - 1 \}$ and for all $x, y \in \mathbb{Z}_n$ we define:

(1)Consider the generated subgroup $(\langle 1 \rangle, \cdot)$. Note that $1 \cdot 1 = 2$, $(1 \cdot 1) \cdot 1 = 3$, …, $\underbrace{(1 \cdot 1 \cdot ... \cdot 1) \cdot 1}_{n-1 \: \mathrm{many \: factors}} = n - 1$, and $\underbrace{(1 \cdot 1 \cdot ... \cdot 1) \cdot 1}_{n \: \mathrm{many \: factors}} = 0$. Therefore the generated subgroup $\langle 1 \rangle$ contains all of the elements in $\mathbb{Z}_n$, that is:

(2)Therefore we can say that $(\mathbb{Z}_n, +)$ is a cyclic group and $1$ is a generator for this group. In fact, every element in $\mathbb{Z}_n$ is a generator for $(\mathbb{Z}_n, +)$.

Of course, this is not always the case. Now consider the group $(\mathbb{Z}_n, \cdot)$ where $n$ is NOT prime and $\cdot$ is the operation defined for all $x, y \in \mathbb{Z}_n$ by:

(3)Take $n = 4$ for example. Then consider the element $2 \in \mathbb{Z}_n$. We have that:

(4)Hence $\mathbb{Z}_n \neq \langle 2 \rangle$ with the operation $\cdot$ so $2$ is not a generator for $\mathbb{Z}_n$. The reader can verify that the only generators of $\mathbb{Z}_n$ are $1$ and $3$.

For a third example, consider the group $(\mathbb{R}, +)$. This group has NO generator. To show this, suppose that $(\mathbb{R}, +$ does have a generator. Then there exists an $a \in \mathbb{R}$ such that for all $\mathbb{R} = \langle a \rangle$. But $\langle a \rangle$ is a countable set and $\mathbb{R}$ is not - a contradiction. So $(\mathbb{R}, +)$ has no generator.

We now state a very nice result which tells us that every cyclic group is an abelian group.

Proposition 1: Let $(G, \cdot)$ be a cyclic group. Then $(G, \cdot)$ is an abelian group. |

**Proof:**Let $(G, \cdot)$ be a cyclic group. Then there exists an element $a \in G$ such that:

- Let $x, y \in G$. Then $x = a^m$ and $y = a^n$ for some $m, n \in \mathbb{Z}$. Therefore:

- So $x \cdot y = y \cdot x$ for all $x, y \in G$ which means that $(G, \cdot)$ is an abelian group. $\blacksquare$

The converse of Theorem 1 gives us a simple criterion for quickly determining when a group is not a cyclic group in certain circumstances.

Corollary 2: If $(G, \cdot)$ is not an abelian group then $(G, \cdot)$ is not a cyclic group. |